For the function given​ below, find a formula for the Riemann sum obtained by dividing the interval ​[00​,22​] into n equal subintervals and using the​ right-hand endpoint for each c Subscript kck. Then take a limit of this sum as n right arrow infinityn → ∞ to calculate the area under the curve over ​[00​,22​].f left parenthesis x right parenthesis equals x squared plus 3f(x)=x2+3Write a formula for a Riemann sum for the function f left parenthesis x right parenthesis equals x squared plus 3f(x)=x2+3 over the interval ​[00​,22​].

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Asked Dec 2, 2019
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For the function given​ below, find a formula for the Riemann sum obtained by dividing the interval ​[
0
0​,
2
2​] into n equal subintervals and using the​ right-hand endpoint for each
c Subscript k
ck. Then take a limit of this sum as
n right arrow infinity
n → ∞ to calculate the area under the curve over ​[
0
0​,
2
2​].

f left parenthesis x right parenthesis equals x squared plus 3
f(x)=x2+3
Write a formula for a Riemann sum for the function
f left parenthesis x right parenthesis equals x squared plus 3
f(x)=x2+3 over the interval ​[
0
0​,
2
2​].

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Expert Answer

Step 1

Given the interval [0,2] is to be divided into n subintervals and right-hand endpoint for each interval is to be used. The function is:

 

Dividing the interval into subintervals:

 

So the right end points are given by:

The Riemann Sum for a function is given by:

Where x1, x2, x3… are the endpoints considered.

f(x) x +3
2-0 2
п
п
2 2 2 2
2 2
+-+
(n terms)
пп
пп
п
246
2
_
ппп
= Arf (x)+Axf (x, ) + Axf (x, ).+ Axf (x,)
+
help_outline

Image Transcriptionclose

f(x) x +3 2-0 2 п п 2 2 2 2 2 2 +-+ (n terms) пп пп п 246 2 _ ппп = Arf (x)+Axf (x, ) + Axf (x, ).+ Axf (x,) +

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Step 2

Using the above formula, the R...

-EME)-€M€}+€M€J<r
A =
n
+3 +
(2)
+3
A =
+3
+3
A
+3
+3
n
+(2) +(3) (n)
A
+..+
n
43
(2n)
3n
A =
3
n
(2.2)
(2-n)
+3n
(2.3)
(2.1)
A
..+
n
-P-)
2.13
2 39
23
.33
2
A
+3n
..+
23
(P +2 +3++3
A =
-fE)
8 n(n+1)
A =
+37
4
n
n(n+1)
(3n)
n'
n
4(n+1)
+6
A =
help_outline

Image Transcriptionclose

-EME)-€M€}+€M€J<r A = n +3 + (2) +3 A = +3 +3 A +3 +3 n +(2) +(3) (n) A +..+ n 43 (2n) 3n A = 3 n (2.2) (2-n) +3n (2.3) (2.1) A ..+ n -P-) 2.13 2 39 23 .33 2 A +3n ..+ 23 (P +2 +3++3 A = -fE) 8 n(n+1) A = +37 4 n n(n+1) (3n) n' n 4(n+1) +6 A =

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Math

Calculus

Integration